\newproblem{lay:7_1_35}{
  % Problem identification
	\begin{large}
	  \hspace{\fill}\newline
    \textbf{Lay, 7.1.35}
	\end{large}
	\\
  \ifthenelse{\boolean{identifyAuthor}}{\textit{Carlos Oscar Sorzano, Aug. 31st, 2013} \\}{}

  % Problem statement
	Let $\mathbf{u}$ be a unit vector in $\mathbb{R}^n$, and let $B=\mathbf{u}\mathbf{u}^T$.
	\begin{enumerate}[a.]
		\item Given any $\mathbf{x}\in\mathbb{R}^n$, compute $B\mathbf{x}$ and show that $B\mathbf{x}$ is the orthogonal projection of $\mathbf{x}$ onto $\mathbf{u}$,
		      as described in Section 6.2.
		\item Show that $B$ is a symmetric matrix and $B^2=B$.
		\item Show that $\mathbf{u}$ is an eigenvector of $B$. What is the corresponding eigenvalue?
	\end{enumerate}
}{
   % Solution
	\begin{enumerate}[a.]
		\item The orthogonal projection of $\mathbf{x}$ onto $\mathbf{u}$ is defined as
		      \begin{center}
						$\mathrm{Proj}_{\mathbf{u}}\{\mathbf{x}\}=\frac{\mathbf{x}\cdot\mathbf{u}}{\|\mathbf{u}\|^2}\mathbf{u}$
					\end{center}
					Since $\mathbf{u}$ is unitary, $\|\mathbf{u}\|^2=1$, then
		      \begin{center}
						$\begin{array}{rcll}
						   \mathrm{Proj}_{\mathbf{u}}\{\mathbf{x}\}&=&(\mathbf{x}\cdot\mathbf{u})\mathbf{u}& [\text{inner product is commutative}] \\
							    &=&(\mathbf{u}\cdot\mathbf{x})\mathbf{u} & [\text{by definition of inner product}] \\
									&=&(\mathbf{u}^T\mathbf{x})\mathbf{u} & [\text{$\mathbf{u}^T\mathbf{x}$ is a scalar}] \\
								  &=&\mathbf{u}(\mathbf{u}^T\mathbf{x}) & [\text{associativity of matrix multiplication}] \\
									&=&(\mathbf{u}\mathbf{u}^T)\mathbf{x} & [B=\mathbf{u}\mathbf{u}^T] \\
									&=&B\mathbf{x} & \\
						\end{array}$
					\end{center}
		\item $B$ is symmetric as shown in Exercise 7.1.27. Let's show now that $B^2=B$
		      \begin{center}
						$B^2=(\mathbf{u}\mathbf{u}^T)(\mathbf{u}\mathbf{u}^T)=\mathbf{u}(\mathbf{u}^T\mathbf{u})\mathbf{u}^T$
					\end{center}
					But $\mathbf{u}^T\mathbf{u}=1$ because $\mathbf{u}$ is unitary. Then,
		      \begin{center}
						$B^2=\mathbf{u}\mathbf{u}^T=B$
					\end{center}
					The fact that $B^2=B$ implies that projecting $\mathrm{Proj}_{\mathbf{u}}\{\mathbf{x}\}$ onto $\mathbf{u}$ (applying the projection operation twice) has no
					effect.
		\item Let's calculate the product $B\mathbf{u}$
		      \begin{center}
						$B\mathbf{u}=(\mathbf{u}\mathbf{u}^T)\mathbf{u}=\mathbf{u}(\mathbf{u}^T\mathbf{u})=\mathbf{u}$
					\end{center}
					So, its eigenvalue is 1. The meaning of this latter property is that the orthogonal projection of $\mathbf{u}$ onto $\mathbf{u}$ is $\mathbf{u}$ itself.
		
	\end{enumerate}
}
\useproblem{lay:7_1_35}
\ifthenelse{\boolean{eachProblemInOnePage}}{\newpage}{}

